Torque
5-5. If the force F in Fig. 5-11a is equal to 80 lb, what is the resultant torque about axis A neglecting the weight of the rod. What is the resultant torque about axis B? Counterclockwise torques are positive, so that τ (a) τ
A A

is - and τ

B

is +.

= (80 lb)(0.845 ft) = -67.6 lb ft

(b) τ

B

= (80 lb)(1.27 ft) = +101 lb ft

5-6. The force F in Fig. 5-11b is 400 N and the angle iron is of negligible weight. What is the resultant torque about axis A and about axis B? Counterclockwise torques are positive, so that τ (a) τ
A A

is + and τ

B

is -.

= (400 N)(1.732 m) = +693 N m;

(b) τ

B

= (400 N)(2.50 m) = -1000 N m

36

Chapter 5 Torque and Rotational Equilibrium

Physics, 6th Edition

5-7. A leather belt is wrapped around a pulley 20 cm in diameter. A force of 60 N is applied to the belt. What is the torque at the center of the shaft? r = ½D = 10 cm; τ = (60 N)(0.10 m) = +6.00 N m F

5-9. A person who weighs 650 N rides a bicycle. The pedals move in a circle of radius 40 cm. If the entire weight acts on each downward moving pedal, what is the maximum torque? τ = (250 N)(0.40 m) τ = 260 N m

5-10. A single belt is wrapped around two pulleys. The drive pulley has a diameter of 10 cm, and the output pulley has a diameter of 20 cm. If the top belt tension is essentially 50 N at the edge of each pulley, what are the input and output torques? Input torque = (50 N)(0.10 m) = 5 N m Output torque = (50 N)(0.20 m) = 10 N m

Equilibrium
5-19. A uniform meter stick is balanced at its midpoint with a single support. A 60-N weight is suspended at the 30 cm mark. At what point must a 40-N weight be hung to balance the system? (The 60-N weight is 20 cm from the axis)
60 N 20 cm

The 5-N weight must be placed at the 90-cm mark 5-21. An 8-m board of negligible weight is supported at a point 2 m from the right end where a 50-N weight is attached. What downward force at the must be exerted at the left end to produce equilibrium? Σ τ = 0: F (6 m) – (50 N)(2 m) = 0 or F = 16.7 N F

6m

2m

6 F = 100 N m

50 N

39

Chapter 5 Torque and Rotational Equilibrium

Physics, 6th Edition

5-22. A 4-m pole is supported at each end by hunters carrying an 800-N deer which is hung at a point 1.5 m from the left end. What are the upward forces required by each hunter? Σ τ = A (0) – (800 N)(1.5 m) + B (4.0 m) = 0 4B = 1200 N or B = 300 N A = 500 N A 1.5 m
Axis

5-27. A bridge whose total weight is 4500 N is 20 m long and supported at each end. Find the forces exerted at each end when a 1600-N tractor is located 8 m from the left end. Στ B = 2890 N
Α

= B (20 m) – (1600 N)(8 m) – (4500 N)( 10 A = 0; m)
8m 2m 10 m

B

Σ Fy = A + B – 1600 N – 4500 N = 0 B = 3210 N

Axis 1600 N 4500 N

A = 6100 N – B = 6100 N – 2890 N;

5-28. A 10-ft platform weighing 40 lb is supported at each end by stepladders. A 180-lb painter is located 4 ft from the right end. Find the forces exerted by the supports. Στ B = 128 lb
Α

= B(10 ft) – (40 lb)(5 ft) – (180 lb)( 6 ft) = 0; A
Axis

B
5 ft 1 ft 4 ft

Σ Fy = A + B – 40 lb – 180 lb = 0 A = 92.0 lb

A = 220 lb – B = 220 lb – 128 lb;

40 lb

180 lb

*5-29. A horizontal, 6-m boom weighing 400 N is hinged at the wall as shown in Fig. 5-19. A cable is attached at a point 4.5 m away from the wall, and a 1200-N weight is attached to the right end. What is the tension in the cable? φ = 90 – 37 = 53 ; Ty = T sin 53
0 0 0 0

V
3m Axis

Ty Ty 1.5 m 400 N

B

1.5 m

H

Στ

Α

= (T sin 530)(4.5 m) – (400 N)(3 m) – (1200 N)(6 m) = 0; T = 2340 N

1200 N

3.59 T = 1200 N + 7200 N;

*5-30. What are the horizontal and vertical components of the force exerted by the wall on the boom? What is the magnitude and direction of this force? Σ Fx = H – Tx = 0; H – T cos 530 = 0; H = (2340 N) cos 530; H = 1408 N

Center of Gravity
5-31. A uniform 6-m bar has a length of 6 m and weighs 30 N. A 50-N weight is hung from the left end and a 20-N force is hung at the right end. How far from the left end will a single upward force produce equilibrium? Σ Fy = F – 50 N – 30 N – 20 N = 0; F = 100 N
Axis

The forces exerted by the supports are : A = 375 N and B = 425 N 5-38. An 8-m steel metal beam weighs 2400 N and is supported 3 m from the right end. If a 9000-N weight is placed on the right end, what force must be exerted at the left end to balance the system? Στ
4m
Α

Critical Thinking Questions
*5-41. A 30-lb box and a 50-lb box are on opposite ends of a 16-ft board supported only at its midpoint. How far from the left end should a 40-lb box be placed to produce equilibrium? Would the result be different if the board weighed 90 lb? Why, or why not? Στ = (30 lb)(8 ft) + (40 lb)(x) – (50 lb)(8 ft) = 0; x 30 lb 40 lb 8 ft F 8 ft W 50 lb

x = 4.00 ft Note that the weight acting at the center of the board does NOT contribute to torque about

the center, and therefore, the balance point is not affected, regardless of the weight. 5-42. On a lab bench you have a small rock, a 4-N meterstick and a single knife-edge support. Explain how you can use these three items to find the weight of the small rock. a 44 b

Chapter 5 Torque and Rotational Equilibrium Measure distances a and b; determine F and then F calculate the weight W from equilibrium methods.

Physics, 6th Edition
0.5 m

4N

W

*5-43. Find the forces F1, F2, and F3 such that the system drawn in Fig. 5-23 is in equilibrium. Note action-reaction forces R and R’. First, lets work with top board: Σ τ (about R) = 0; Force R is upward. Στ
R

F1
2 ft 6 ft

R
2 ft

50 lb

F2
3 ft 300 lb 5 ft 200 lb

F3
2 ft

= (300 lb)(6 ft) – (50 lb)(2 ft) – F1(8 ft) = 0

R’ ’

F1 = 213 lb

Now, Σ Fy = 0 gives: 213 lb + R –300 lb – 50 lb = 0; R = 138 lb = R’

Next we sum torques about F2 with R’ = 138 lb is directed in a downward direction: Στ
F

*5-49. An car has a distance of 3.4 m between front and rear axles. If 60 percent of the weight rests on the front wheels, how far is the center of gravity located from the front axle? F Axis x Σ τ = 0.6W(0) + 0.4W(3.4 m) – F x = 0 0.4W
3.4 m